 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-3/5d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Every

b^{m}

lies in

N

because

(b N)^{m} = N

. Conversely choose

r

with

m r \equiv 1 (mod n)

; for

a \in N

a = (a^{r})^{m}

because

a^{n} = e

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